12 Days of Christmas GMAT Competition - Day 2: On a San Diego Comic-Co

Question

12 Days of Christmas GMAT Competition with Lots of Fun

On a San Diego Comic-Con, Sheldon bought all 72 identical collectible Mr. Spock action figures. He paid a total of $a6,79b where a is the ten thousands digit and b is the units digit. If each figure cost the same and the cost in dollars was an integer, what is the value of a + b ?

A. 4
B. 5
C. 6
D. 9
E. 10

M37-74

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Official Answer

B

Bunuel · Accepted Answer

Official Solution:

On a San Diego Comic-Con, Sheldon bought all 72 identical collectible Mr. Spock action figures. He paid a total of  $a6,79b  where  a  is the ten thousands digit and  b  is the units digit. If each figure cost the same and the cost in dollars was an integer, what is the value of  a + b  ?

A.  4 
B.  5 
C.  6 
D.  9 
E.  10

To solve this question, we should know divisibility rule for 8 and 9.
 
 Divisibility rule for  2^n : 
 
An integer is divisible by 2 if its last digit is divisible by 2.
 
An integer is divisible by 4 if its last two digits is divisible by 4.
 
An integer is divisible by 8 if its last three digits is divisible by 8.
 
...
 
Generally, an integer is divisible by  2^n  if its last  n  digits is divisible by  2^n . ...
 
 Divisibility rule for  9 : 
 
The rule for divisibility by 9 is similar to divisibility rule for 3. That is, if the sum of digits of the number is divisible by 9, then the number itself is divisible by 9.
 
 Back to the question: 
 
We are given that  72*p=a6,79b , where  p  is an integer price of each figure.
 
 p=\frac{a6,79b}{8*9}=integer . So,  a6,79b  must be divisible by both 8 and 9.
 
According to the rule above,  a6,79b  will be divisible by 8, whenever  79b  is divisible by 8. 800 is divisible by 8, next smallest number divisible by 8 is 792, so  b=2 
 
 a6,79b  will be divisible by 9, if  a+6+7+9+b=a+6+7+9+2=24+a  is divisible by 9, so when  a=3 .
 
Hence,  a+b=2+3=5 .

Answer: B

vivek1408 · Answer

If a679b is perfectly divisible by 72, then it is divisible by 4,8,3 and 9. Applying divisibility rules for each:
1. Div by 4 only when b=2 or 6 ( last two digits divisible by 4 ie. 9b%4=0)
2. Doc by 8 only when b=2 ( last three digits divisible by 8 ie. 79b%4=0)

Now the number is a6792
3. Divisibility rule by 9: sum of digits is divisible by 9 (a+24 is multiple of 9) therefore a=3

Finally number is 36792
a+b = 5

Posted from my mobile device

Deepakjhamb · Answer

On a San Diego Comic-Con, Sheldon bought all 72 identical collectible Mr. Spock action figures. He paid a total of $a6,79b where a is the ten thousands digit and b is the units digit. If each figure cost the same and the cost in dollars was an integer, what is the value of a + b ?

A. 4
B. 5
C. 6
D. 9
E. 10

sol : let price of each action figure be x

then 72 * x = a,679b

so here x is an integer , therfore a679b needs to be divisible by 72 or we can say that divisible by 9 and 8 both

now apply divisibility rule of 8 , we need to have last 3 digits divisible by 8 so options we have for be are even numbers lets say 0,2,4,6 ,8 we see that only 2 satisfies this requirement so b=2

now lets apply divisibility rule for 9 , a number is divisible by 9 if sum of digits is divisible by 9

we get 6+7+9+ 2( value of b we found out) + a has to be divisible by 9

so we get 24 + a divisible by 9 so we get a=3 as the only value possible as is a digit

so a+ b = 3+2 = 5

so the answer is B) 5

197369 · Answer

solution :
72 * (cost of each figure) = a6,79b
since each cost is identical , so 72 should divide the amount wholly i.e. 72 is a factor of the amount. 
Also, 72 = 8*9 (this means the amount should be divisible by 9)
thus, for any number to be divisible by 9, the sum of the digits must be divisible by 9.
a+6+7+9+b = (a+b+22) should be divisible by 9 
Thus, the nearest number is 27. so (a+b) = 27-22=5

option B is the answer.

RohitSaluja · Answer

The correct answer is B, below is the explanation.

The question states that each figure costs the same and the total cost was an integer, hence a679b should be divisible by 2 and 3(which are the only prime factor of 72), for a number to be divisible by 3, the sum of all digits should be divisible by 3. summing a679b, we get a + b + 22, Now plugging in the options

A. 4 - This would make the sum 26 which is not divisible by 3, so not the answer
B. 5- This would make the sum 27 which is divisible by 3, hence this is the correct answer
C. 6 - This would make the sum 28 which is not divisible by 3, so not the answer
D. 9 - This would make the sum 31 which is not divisible by 3, so not the answer
E. 10 - This would make the sum 32 which is not divisible by 3, so not the answer

QuantMadeEasy · Answer

In order to divide $a6,79b into 72 equal parts, the number $a6,79b to be divisible by 8 and 9 both.
 Divisibility rule for 8 of last three digits to be divisible by 8 
therefore 79b is divisible by 8 when b = 2

Divisibility rule for 9 is sum of all the digits to be divisible by 9 
therefore a+6+7+9++2 is divisible by 9 when a = 3

Hence a+b = 5

IMO b

ashu2526 · Answer

a679b/72

72 is a multiple of 9
for a number to be completely divisible it must be a multiple of 9 
Digital Sum(a+6+7+9+b) = 4
So according to this a+b = 5
Which is an option but lets confirm it 
possibility a/b (2,3) (0,5) (4,1)
72 is even so b should be even possible b = 0,2,4 and a = 3, 5, 1

Now some solving (Hit and Trail) a = 3 and b = 2 satisfy

36792/72 = 511

Answer 5 (B)

Marcost · Answer

For a number to be divided by 72, it must be divided by 8 and 9.

To know if a number is divisible by 8, take last 3 digits and prove if that number is dividied by 8.
To know if a number is divisible 9, the sum of its digits must be divisible by 9.

b must be 2 so that 79b is divisible by 8. 
a+6+7+9+2 = 9k then, a must be 3

a+b = 3+2 = 5

plaverbach · Answer

price= \frac{a6,79b}{72} 
and
72=2.2.2.3.3
Therefore: 
 a+6+7+9+b  is  divisible by 3

Since 6 and 9 are divisible by 3, we've got to make 7 be divisible as well
7+b+a needs to be divisible by 3

lets plug in the answers:
A. 4: 7+4=  11 is NOT divisible by 3 
 B.  5: 7+5=   12 is divisible!  
C. 6: 7+6=  13 is NOT divisible by 3 
D. 9: 7+9=  16 is NOT divisible by 3 
E. 10: 7+10=  17 is NOT divisible by 3

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